Problem: Kana was just an ordinary high school girl before a talent scout discovered her. Then, she became an idol. But different from the stereotype, she is also a gameholic. 

One day Kana gets interested in a new adventure game called Dragon Quest. In this game, her quest is to beat a dragon.

<image>

The dragon has a hit point of x initially. When its hit point goes to 0 or under 0, it will be defeated. In order to defeat the dragon, Kana can cast the two following types of spells. 

  * Void Absorption

Assume that the dragon's current hit point is h, after casting this spell its hit point will become \left⌊ h/2 \right⌋ + 10. Here \left⌊ h/2 \right⌋ denotes h divided by two, rounded down.

  * Lightning Strike

This spell will decrease the dragon's hit point by 10. Assume that the dragon's current hit point is h, after casting this spell its hit point will be lowered to h-10.




Due to some reasons Kana can only cast no more than n Void Absorptions and m Lightning Strikes. She can cast the spells in any order and doesn't have to cast all the spells. Kana isn't good at math, so you are going to help her to find out whether it is possible to defeat the dragon.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The next t lines describe test cases. For each test case the only line contains three integers x, n, m (1≤ x ≤ 10^5, 0≤ n,m≤30) — the dragon's intitial hit point, the maximum number of Void Absorptions and Lightning Strikes Kana can cast respectively.

Output

If it is possible to defeat the dragon, print "YES" (without quotes). Otherwise, print "NO" (without quotes).

You can print each letter in any case (upper or lower).

Example

Input


7
100 3 4
189 3 4
64 2 3
63 2 3
30 27 7
10 9 1
69117 21 2


Output


YES
NO
NO
YES
YES
YES
YES

Note

One possible casting sequence of the first test case is shown below:

  * Void Absorption \left⌊ 100/2 \right⌋ + 10=60.
  * Lightning Strike 60-10=50.
  * Void Absorption \left⌊ 50/2 \right⌋ + 10=35.
  * Void Absorption \left⌊ 35/2 \right⌋ + 10=27.
  * Lightning Strike 27-10=17.
  * Lightning Strike 17-10=7.
  * Lightning Strike 7-10=-3.

Solution (in PYTHON):

Solution:
from math import factorial as fac
from collections import defaultdict
from copy import deepcopy
import sys, math
f = None
try:
	f = open('q1.input', 'r')
except IOError:
	f = sys.stdin
if 'xrange' in dir(__builtins__):
	range = xrange
# print(f.readline())
# sys.setrecursionlimit(10**6)


def print_case_iterable(case_num, iterable):
	print("Case #{}: {}".format(case_num," ".join(map(str,iterable))))

def print_case_number(case_num, iterable):
	print("Case #{}: {}".format(case_num,iterable))

def print_iterable(A):
	print (' '.join(A))

def read_int():
	return int(f.readline())
def read_int_array():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def read_string():
	return list(f.readline().strip())

NUMBER = 10**9 + 7
# NUMBER = 998244353
def factorial(n) : 
	M = NUMBER
	f = 1
  
	for i in range(1, n + 1):  
		f = (f * i) % M # Now f never can  
						# exceed 10^9+7  
	return f
def mult(a,b):
	return (a * b) % NUMBER

def minus(a , b):
	return (a - b) % NUMBER

def plus(a , b):
	return (a + b) % NUMBER

def egcd(a, b):
	if a == 0:
		return (b, 0, 1)
	else:
		g, y, x = egcd(b % a, a)
		return (g, x - (b // a) * y, y)

def modinv(a):
	m = NUMBER
	g, x, y = egcd(a, m)
	if g != 1:
		raise Exception('modular inverse does not exist')
	else:
		return x % m
from collections import deque, defaultdict 
import heapq

def bfs(V, E):
	s = set()
	print('\n'.join(map(str,E.items())))
	cnt = 0 
	for v in V:
		if v in s:
			continue
		q = deque([])
		s.add(v)
		q.append(v)
		while len(q) > 0:
			x = q.popleft()
			for e in E[x]:
				if e not in s:
					s.add(e)
					q.append(e)
			print(q)
		cnt += 1
	return cnt
		
# from typing import List, Tuple

def binary_count_smaller(L, x):
	l = 0
	if x < L[0]:
		return 0
	r = len(L)
	while r - l > 1:
		mid = (l+r)//2
		if L[mid] < x:
			l = mid
		else:
			r = mid
	return l + 1
def binary_count_larger(L, x):
	l = 0
	if x > L[-1]:
		return 0
	r = len(L)
	while r - l > 1:
		mid = (l+r)//2
		if L[mid] <= x:
			l = mid
		else:
			r = mid
	return len(L) - (r-1)

def regular_bs(L,x,l=None,r=None):
	l = 0 if l == None else l
	r = len(L) if r == None else r
	while (r-l) > 1:
		m = (r+l)//2
		l = m if L[m] <= x else l
		r = m if L[m] > x else r
	cnt = 0
	return l

from itertools import product

def preprocess():
	nums = dict(zip(sequences,range(10)))
	changes = defaultdict(dict)
	for comb in map(lambda x: ''.join(x), product(*([['0','1']]*7))):
		for seq in sequences:
			a = int(comb,2)
			b = int(seq,2)
			num_changes = bin(a^b).count('1')
			if num_changes == seq.count('1') - comb.count('1'):
				changes[comb][num_changes] = nums[seq]
	return changes


# import StringIO use for PYPY when lots of prints
#example usage:
#	 for l in res:
#	 	print >>output, str(len(l)) + ' ' +  ' '.join(l)
from collections import deque

def solution(x,n,m):
	i = 0
	while x >=20 and i < n:
		x = x//2 + 10
		i+=1
	return 'YES' if x <= 10*m else 'NO'


def main():
	T = read_int()
	for i in range(T):
		x,n,m = read_int_array()
		print(solution(x,n,m))


if __name__ == '__main__':
	main()
# print(regular_bs([1,2],1))